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RING STRUCTURES CONCERNING FACTORIZATION
MODULO RADICALS
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Bull. Korean Math. Soc. 0 (0), No. 0, pp. 1–0
https://doi.org/10.4134/BKMS.b150862
pISSN: 1015-8634 / eISSN: 2234-3016
Hai-Lan Jin, Hong Kee Kim, and Yang Lee
fP
Abstract. The aim in this note is to describe some classes of rings in
relation to factorization by prime radical, upper nilradical, and Jacobson
radical. We introduce the concepts of tpr ring, tunr ring, and tjr ring
in the process, respectively. Their ring theoretical structures are investigated in relation to various sorts of factor rings and extensions. We
also study the structure of noncommutative tpr (tunr, tjr) rings of minimal order, which can be a base of constructing examples of various ring
structures. Various sorts of structures of known examples are studied in
relation with the topics of this note.
1. Structure of tpr rings
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Throughout this note every ring is an associative ring with identity unless
otherwise stated. Let R be a ring. The polynomial (resp., power series) ring
with an indeterminate x over R is denoted by R[x] (resp., R[[x]]) and for any
polynomial (resp., power series) f (x) in R[x] (resp., R[[x]]), let Cf (x) denote
the set of all coefficients of f (x). Use the notation that R̄ = R/I and r̄ = r + I,
where I is an ideal of R. Z (Zn ) denotes the ring of integers (modulo n).
Denote the n by n full (resp., upper triangular) matrix ring over R by M atn (R)
(resp., Un (R)). Use Eij for the matrix with (i, j)-entry 1 and zeros elsewhere.
Following the literature, Dn (R) = {(aij ) ∈ Un (R) | a11 = · · · = ann } and
Nn (R) = {(bij ) ∈ Dn (R) | b11 = · · · = bnn = 0}.
Use I(R) for the set of all nontrivial (i.e., nonzero nonunit) idempotents of
R, and set Ie (R) = I(R) ∪ {0, 1}. A nilpotent element is also said to be a
nilpotent for simplicity. Let J(R), N0 (R), N∗ (R), N ∗ (R), and N (R) to denote
the Jacobson radical, the Wedderburn radical (i.e., sum of all nilpotent ideals),
the prime radical, the upper nilradical (i.e., sum of all nil ideals), and the set
of all nilpotents in R (possibly without identity), respectively. It is well-known
that N ∗ (R) ⊆ J(R) and N0 (R) ⊆ N∗ (R) ⊆ N ∗ (R) ⊆ N (R). A ring R is
usually called semiprimitive if J(R) = 0.
Received October 22, 2015; Revised June 18, 2016.
2010 Mathematics Subject Classification. 16N40, 16S36.
Key words and phrases. tpr ring, tunr ring, tjr ring, polynomial ring, factor ring, noncommutative ring of minimal order, nilradical, Jacobson radical.
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2017
Korean Mathematical Society
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H.-L. JIN, H. K. KIM, AND Y. LEE
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Let R be a ring (not necessarily with identity) and consider the condition
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(∗) ai ∈ N∗ (R) or bi ∈ N∗ (R) for each i whenever f (x)g(x) ∈ N∗ (R)[x]
Pm
Pn
for f (x) = i=0 ai xi , g(x) = j=0 bj xj ∈ R[x], where we let m = n by using
zero coefficients if necessary. Amitsur in [2, Theorem 3] prove that N∗ (R[x]) =
N∗ (R)[x] for any ring R.
It is easy to find examples which do not satisfy the condition (∗) (e.g.,
f (x) = (1, 0) and g(x) = (0, 1) in (Z ⊕ Z)[x], noting N∗ (Z ⊕ Z) = 0). But every
ring does not satisfy the condition (∗), so the following definition makes sense.
Definition 1.1. A ring R (not necessarily with identity) is called trivializable
over prime radical (simply, tpr) if it satisfies the condition (∗).
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The name of tpr is originated from Cohn [8]. Every domain R is clearly tpr
by the fact that f g = 0 for f, g ∈ R[x], implies f = 0 or g = 0. We provide tpr
rings which are not domains as follows. A ring is usually said to be reduced if
it has no nonzero nilpotents. Following the literature, a ring is called Abelian
if every idempotent is central. Reduced rings are easily shown to be Abelian.
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Lemma 1.2. (1) A ring R is tpr if and only if R/N∗ (R) is a domain.
(2) Let R be a domain and E0 = Dm (R) for m ≥ 1. Then Dn (E0 ) is a tpr
ring for all n ≥ 1.
(3) Let R be a tpr ring. Then Ie (R) = {0, 1}, i.e., I(R) is empty.
(4) If R is a tpr ring, then N (R) = N∗ (R) = N ∗ (R).
(5) Let R be a tpr ring. Then R is semiprime if and only if R is reduced if
and only if R is a domain.
(6) The class of tpr rings is closed under subrings.
Proof. (1) Let R be a ring and ab ∈ N∗ (R) for a, b ∈ R. Then a ∈ N∗ (R)
or b ∈ N∗ (R) because R is tpr. The sufficiency comes from the fact that
R[x]
R
∼
N∗ (R) [x] = N∗ (R)[x] . Indeed, suppose that R/N∗ (R) is a domain and let
Pm
Pm
f (x)g(x) ∈ N∗ (R)[x] for f (x) = i=0 ai xi , g(x) = j=0 bj xj ∈ R[x]. Then
f (x) ∈ N∗ (R)[x] or g(x) ∈ N∗ (R)[x] because N∗R[x]
(R)[x] is a domain.
(2) Let E0 = Dm (R) and E = Dn (E0 ) for m, n ≥ 1. Then
N∗ (E) = {(aij ) ∈ E | aii ∈ Nm (R) for all i},
so we have E/N∗ (E) ∼
= R is a domain. Thus E is a tpr ring by (1).
(3) Let e ∈ Ie (R) and consider the equality e(1 − e) = 0. Since R is tpr,
e ∈ N∗ (R) or 1 − e ∈ N∗ (R). Thus e = 0 or e = 1.
(4) is shown by (1).
(5) is an immediate consequence of (1) and (4), noting that reduced rings
are semiprime.
(6) Let R be a tpr ring and S be a subring of R. Then N (R) = N∗ (R)
by (4), so we get N∗ (S) = N∗ (R) ∩ S from N∗ (R) ∩ S ⊆ N∗ (S) ⊆ N (S) =
N (R) ∩ S = N∗ (R) ∩ S.
RING STRUCTURES CONCERNING FACTORIZATION MODULO RADICALS
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Pm
Pn
Let f (x)g(x) ∈ N∗ (S)[x] for f (x) = i=0 ai xi , g(x) = j=0 bj xj ∈ S[x].
Then f (x)g(x) ∈ N∗ (R)[x]. But, since R is tpr, ai ∈ N∗ (R) or bi ∈ N∗ (R).
This implies that ai ∈ N∗ (S) or bi ∈ N∗ (S) because N∗ (S) = N∗ (R) ∩ S. Thus
S is tpr.
Lemma 1.2(1) implies that R/N∗ (R) is a reduced ring when a given ring R is
tpr. Every tpr ring is Abelian by Lemma 1.2(3). It is possible to give examples
of Abelian rings but not tpr, by help of [17, Lemma 2]. Recall that for each
 f 0 0 ··· 0 
0 f 0 ··· 0
 0 0 f ··· 0 
e ∈ Ie (Dn (R)) is of the form  . . .
 with f ∈ Ie (R) by [17, Lemma 2],
.. .. .. . . . ...
0 0 0 ··· f
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where R is an Abelian ring. So if R is a domain, Ie (Dn (R)) = {0, 1}, noting
that R is tpr by Lemma 1.2(1).
 3 0 0 ··· 0 
0 3 0 ··· 0
Let next R = Z6 .
··· 0 
0 4 0 ··· 0
 0 0 4 ··· 0 
4 0 0
0 0 3 ··· 0
Then 32 = 3, 42 = 4 ∈ I(Z6 ) and  . . . . . ,
.. .. .. . . ..
0 0 0 ··· 3
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.. .. .. . . .. ∈ I(Dn (R)) when n ≥ 2. This implies that Dn (Z6 ) is not tpr by
... ..
0 0 0 ··· 4
Lemma 1.2(3), but Abelian by [17, Lemma 2].
The converse of Lemma 1.2(4) need not hold too. Let R = U2 (A) over a
domain A. Then N (R) = N∗ (R) clearly, but I(R) contains E11 ; hence R is not
tpr by Lemma 1.2(3).
Considering Lemma 1.2(1), one may ask whether a ring R is tpr when
R/N ∗ (R) is a domain. But the answer is negative by the following.
Example 1.3. (1) We refer the ring constructions and computations in [19,
Example 1.2] and [20, Theorem 2.2]. Let A be a domain and Rn = D2n (A),
where n ≥ 1. Define a map
M 0
σ : Rn → Rn+1 by M 7→
.
0 M
Then Rn can be considered as a subring of Rn+1 via σ (i.e., M = σ(M ) for
M ∈ Rn ). Set R = lim Rn be the direct limit of {Rn , σnm }, noting that
−→
σnm = σ m−n (when n ≤ m) is a direct system over I = {1, 2, . . .}. It is easily
checked that N ∗ (R) = {(aij ) ∈ R | aii = 0 for all i}, so R/N ∗ (R) ∼
= A.
But N∗ (R) = 0 by [20, Theorem 2.2], entailing N∗ (R) ( N ∗ (R). Thus R is
2
not tpr by Lemma 1.2(4). Indeed, E12
= 0 but E12 ∈
/ N∗ (R).
(2) There exists a tpr ring whose structure is similar to the ring R in (1).
Let B be a ring with a nonzero nilpotent ideal K such that B/K is a domain
(e.g., B = Z4 with K = 2Z4 ), and
Rn = {(aij ) ∈ D2n (B) | ast ∈ K for all s, t with s < t},
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H.-L. JIN, H. K. KIM, AND Y. LEE
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N0 (R) = N∗ (R) = N ∗ (R) = N (R) = {(aij ) ∈ R | aii ∈ K for all i},
t
0
lim Rn
where n ≥ 1. Define a map σ : Rn → Rn+1 by M 7→ ( M
0 M ) and let R = −
→
be the direct limit of {Rn , σnm } as in (1). Then we have
m
entailing R/N ∗ (R) ∼
= B/K. In fact, N0 (R)m = 0 when K m = 0. Thus R is
tpr by Lemma 1.2(1).
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Note that the fact of N∗ (R) = N (R), in the proof of Lemma 1.2(6), need
not hold for any ring R. Indeed, consider the ring R in Example 1.3(1), 0 =
N∗ (R) ( N ∗ (R) and N (R) = N ∗ (Rn ) = N ∗ (D2n (A)) = {(aij ) ∈ D2n (A) |
aii = 0 for all i} =
6 0, where Rn is a subring of R.
Armendariz [5, Lemma 1] proved that, for a reduced ring R,
ab = 0 for all a ∈ Cf (x) , b ∈ Cg(x) whenever f (x)g(x) = 0
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where f (x), g(x) ∈ R[x]. Rege et al. [30] called a ring (possibly without identity) Armendariz if it satisfies this property, based on Armendariz’s result. So
reduced rings are clearly Armendariz. Armendariz rings are Abelian by the
proof of [3, Theorem 6] or [18, Corollary 8]. But Armendariz rings need not
be tpr as can be seen the non-tpr ring Z ⊕ Z. Tpr rings also need not be
Armendariz. Indeed, Dn (A), over a domain A, is a tpr ring by Lemma 1.2(2);
but Dn (A) is not Armendariz by [23, Example 3] when n ≥ 4.
Let A be an algebra, with or without identity, over a commutative ring S.
Following Dorroh [9], the Dorroh extension of A by S is the Abelian group
A ⊕ S with multiplication given by (r1 , s1 )(r2 , s2 ) = (r1 r2 + s1 r2 + s2 r1 , s1 s2 )
for ri ∈ A and si ∈ S. This algebraic system is denoted by A ⊕D S.
If a ring R is an Armendariz ring, then N0 (R) = N∗ (R) = N ∗ (R) by [22,
Lemma 2.3(5)]. So it is natural to compare this with the fact that if R is a tpr
ring, then N∗ (R) = N ∗ (R) = N (R) by Lemma 1.2(4). The following makes a
distinction between these two facts.
Example 1.4. (1) There exists an Armendariz ring R such that R is not tpr
and N ∗ (R) ( N (R). Let K be a field and A = Kha, bi be the free algebra with
noncommuting indeterminates a, b over K. Let I be the ideal of A generated
by b2 , and set R = A/I. We identify a, b with their images in R for simplicity.
Then R is Armendariz by [4, Example 4.8]. Let 0 6= s ∈ N (R). Then s is of
the form either kb or btb for some k ∈ K and t ∈ R by the argument in [21,
pages 4–5]. However ba is not a nilpotent, forcing ba ∈
/ N ∗ (R). This implies
∗
N (R) ( N (R), showing that R is not tpr by Lemma 1.2(4).
(2) We refer the construction and computation in [24, Example 2.4(3)] which
extends the argument in Klein [27, Example (ii)]. Let S be the factor ring
of the polynomial ring Z2 [t1 , t2 , . . .] with a set of commuting indeterminates
{ti | i = 1, 2, . . .} over Z2 , modulo the ideal generated by {t2i | i = 1, 2, . . .}.
We identify ti ’s with their images in R for simplicity. It is easily checked
that N0 (S) is the set of all polynomials of zero constants in S. Following [24,
RING STRUCTURES CONCERNING FACTORIZATION MODULO RADICALS
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N0 (S)
S
T =
and R = T ⊕D Z2 .
N0 (S) N0 (S)
t
Example 2.4(3)], set
Then, by the argument
2.4(3)], T ⊕D 0 = N (R) = N∗ (R) =
in [24, Example
N0 (S) N0 (S)
N (R) and N0 (R) = N0 (S) N0 (S) ⊕D 0 ( N∗ (R). So R is not Armendariz
by [22, Lemma 2.3(5)]. Indeed, for f (x) = ( t01 t01 , 0) + ( t02 t02 , 0)x ∈ R[x],
we have f (x)2 = 0 and ( t01 t01 , 0)( t02 t02 , 0) = ( t10t2 t10t2 , 0) 6= 0. But since
R
∼
N∗ (R) = Z2 , R is a tpr ring by Lemma 1.2(1).
∗
fP
Following Han et al. [12], a ring R is called APR if
f (x)g(x) ∈ N∗ (R)[x] implies ab ∈ N∗ (R) for all a ∈ Cf (x) and b ∈ Cg(x)
Ah
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where f (x), g(x) ∈ R[x]. So R is APR if and only if R/N∗ (R) is Armendariz,
entailing that a semiprime ring is APR if and only if it is Armendariz. Obviously tpr rings are APR by Lemma 1.2(1). Armendariz rings are APR by [12,
Theorem 1.4(2)]. Therefore the concept of APR ring is a generalization of both
tpr and Armendariz. Un (A), over a reduced ring A, is neither Armendariz nor
tpr when n ≥ 2, because it is non-Abelian; but it is APR since R/N∗ (R) is
reduced (hence Armendariz).
We argue next the case of N0 (R) = N∗ (R) = N ∗ (R) = N (R), considering
the relation N0 (R) ( N∗ (R) = N ∗ (R) = N (R) in Example 1.4(2). We use
deg f (x) to denote the degree of a given polynomial f (x). Birkenmeier et
al. in [6, Proposition 2.6] prove that if N (R) = N∗ (R) for a ring R, then
N (R[x]) = N∗ (R)[x], entailing N (R[x]) = N∗ (R[x]) = N∗ (R)[x] = N (R)[x].
Following [1, p. 130], a subset of a ring R is said to be locally nilpotent if
its finitely generated subrings are nilpotent. Due to [1, p. 130], the Levitzki
radical of R, written by sσ(R), means the sum of all locally nilpotent ideals of
R. It is well-known that N∗ (R) ⊆ sσ(R) ⊆ N ∗ (R). Amitsur in [2, Theorem 1]
show that J(R[x]) = N [x] for any ring R, where N = J(R[x]) ∩ R is a nil ideal
of R which contains sσ(R).
Let R be a ring. The index of nilpotency of a nilpotent a ∈ R is the least
positive integer k such that ak = 0. The index of nilpotency of a subset S of
R is the supremum of the indices of nilpotency of all nilpotents in S. If such a
supremum is finite, t say, then S is said to be of bounded index t of nilpotency.
Camillo et al. in [7, Corollary 4] prove that N0 (R[x]) = N0 (R)[x] for any ring
R.
Proposition 1.5. (1) A ring R is tpr if and only if so is R[x].
(2) If R is a tpr ring, then
J(R[x]) = N (R[x]) = N ∗ (R[x]) = sσ(R[x])
= N∗ (R[x]) = N∗ (R)[x] = sσ(R)[x] = N ∗ (R)[x] = N (R)[x].
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H.-L. JIN, H. K. KIM, AND Y. LEE
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(3) Let R be a tpr ring of bounded index 2 of nilpotency. Then N0 (R) =
N∗ (R) = N ∗ (R) = N (R).
(4) Let R be a tpr ring of bounded index 2 of nilpotency. Then
J(R[x]) = N (R[x]) = N∗ (R[x]) = N0 (R[x]) = N∗ (R)[x] = N0 (R)[x] = N (R)[x].
Proof. (1) Let R be a tpr ring. Then R/N∗ (R) is a domain by Lemma 1.2(1),
∼ R
entailing that N∗R[x]
(R)[x] = N∗ (R) [x] is also a domain. But N∗ (R[x]) = N∗ (R)[x]
Ah
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do
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by [2, Theorem 3], so N∗R[x]
(R[x]) is a domain. Thus R[x] is tpr by Lemma 1.2(1).
The converse holds by Lemma 1.2(6).
(2) Let R be a tpr ring. Then N (R) = N∗ (R) by Lemma 1.2(4). So we get
moreover N (R[x]) = N∗ (R[x]) = N∗ (R)[x] = N (R)[x] by [2, Theorem 3] and [6,
Proposition 2.6]. This yields N (R[x]) = N ∗ (R[x]) = N∗ (R[x]) = N∗ (R)[x] =
N ∗ (R)[x] = N (R)[x].
While, J(R[x]) = N [x] for some nil ideal N of R by [2, Theorem 1], so
N [x] ⊆ N ∗ (R)[x] implies J(R[x]) = N ∗ (R)[x]. This completes the proof.
(3) Let R be a tpr ring of bounded index 2 of nilpotency and assume on
the contrary that N0 (R) ( N∗ (R), say a ∈ N∗ (R)\N0 (R). Then RaR is a
nilpotent ideal (indeed, (RaR)2 = 0) by Hong et al. [15, Lemma 11] in which
they apply the method of Klein in [26]. For, xRx = 0 by [15, Lemma 11]
because x2 = 0 for all x ∈ RaR. This forces a ∈ N0 (R), a contradiction. Thus
we have N0 (R) = N∗ (R) = N ∗ (R) = N (R) by Lemma 1.2(4) since R is tpr.
(4) is shown by (2), (3), and [7, Corollary 4].
The converse of Proposition 1.5(2) does not hold in general. Indeed, R =
Un (A), over a simple domain A, satisfies the property J(R[x]) = N (R[x]) =
N ∗ (R[x]) = N∗ (R[x]) = N∗ (R)[x] = N ∗ (R)[x] = N (R)[x] by help of [6,
Proposition 2.6], but it is not tpr when n ≥ 2 by Lemma 1.2(3), where
N (R) = {(aij ) ∈ R | aii = 0 for all i}.
One may also ask whether a semiprime tpr ring (hence a domain by Lemma
1.2(1)) is semiprimitive, based on Proposition 1.5(2). However the answer is
negative by the following.
Example 1.6. We refer the ring in [14, Example 3]. Let L be the localization
of Z at the prime ideal pZ, where p is an odd prime. We consider next the
quaternions over L, R say. Then R is clearly a domain (hence tpr), so R[x] is
semiprimitive because J(R[x]) = N [x] for some nil ideal N of R by [2, Theorem
1]. But J(R) = pR, so R is not semiprimitive.
A ring R (possibly without identity) is usually called von Neumann regular
(simply, regular) if for every a ∈ R there exists b ∈ R such that aba = a in [10].
It is easily shown that every regular ring R is semiprimitive (i.e., J(R) = 0).
So we obtain the following.
Proposition 1.7. Let R be a regular ring. Then R is APR if and only if R is
reduced if and only if R is Abelian if and only if R is Armendariz.
RING STRUCTURES CONCERNING FACTORIZATION MODULO RADICALS
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Proof. Use [10, Theorem 3.2] and the fact that R is APR if and only if R/N∗ (R)
is Armendariz.
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However the tpr property need not be equivalent to the conditions in Proposition 1.7, as can be seen by the regular reduced ring Z2 ⊕ Z2 which is not tpr.
Following the literature, a ring R is called π-regular if for each a ∈ R there
exist a positive integer n = n(a) and b ∈ R such that an = an ban . Regular
rings are clearly π-regular, and J(R) is easily shown to be nil for every π-regular
ring R. One may also ask whether π-regular APR rings are Armendariz. But
the answer is negative by the ring R = Dn (S) (n ≥ 4) over a division ring
S. Indeed, R is a π-regular ring because R is local and J(R) = N∗ (R) is
nil. Moreover, since R/N∗ (R) ∼
= S is a division ring, R is APR. But R is not
Armendariz by [23, Example 3].
Proposition 1.8. (1) Let R be a π-regular ring. Then R is tpr if and only if
R is a local ring with J(R) = N∗ (R).
(2) Let R be a regular ring. Then R is tpr if and only if R is a division ring.
Ah
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Proof. (1) Suppose that R is tpr. Then N∗ (R) = N ∗ (R) by Lemma 1.2(4).
But J(R) is nil since R is π-regular, entailing J(R) = N∗ (R). Let a ∈ R\J(R).
Then, since R is π-regular, an = an ban for some b ∈ R and n ≥ 1. But
an 6= 0, so an b and ban are nonzero idempotents. This implies an b = 1 = ban
by Lemma 1.2(3), entailing that a is a unit.
The converse comes from Lemma 1.2(1) because R/N∗ (R) is a division ring.
(2) is an immediate consequence of (1) since regular rings are semiprimitive.
We investigate next the structure of noncommutative tpr rings of minimal
order, which may do roles in constructing various kinds of rings. GF (pn )
denotes the Galois field of order pn . Every ring in the following is of order 16.
Example 1.9. (1) R1 = D3 (Z2 )is a noncommutative
tpr ring by Lemma
0 Z2 Z2
1.2(1) because N∗ (R1 ) = J(R1 ) = 0 0 Z2 and R1 /N∗ (R1 ) ∼
= Z2 .
0 0 0
a b (2) Following Xue [33, Example 2], let R2 =
| a, b ∈ GF (22 ) . Then
0 a2
2
N∗ (R2 ) = J(R2 ) = 00 GF0(2 ) and R2 /N∗ (R2 ) ∼
= GF (22 ). So R2 is a noncommutative tpr ring by Lemma 1.2(1).
(3) Following [33, Example 2], let A1 = Z4 hx, yi be the free algebra with
noncommuting indeterminates x, y over Z4 , and R3 = A1 /I, where I is the
ideal of A1 generated by x3 , y 3 , yx, x2 − xy, x2 − 2, y 2 − 2, 2x, 2y. Identify x and
y with their images in R3 for simplicity. Then N∗ (R3 ) = J(R3 ) = 2Z4 + (x, y),
where (x, y) is the ideal of R3 generated by x and y. So R3 /N∗ (R3 ) ∼
= Z4 , not
reduced, and hence R3 is not a tpr ring.
(4) Following [33, Example 2], let A2 = Z2 hx, yi be the free algebra with
noncommuting indeterminates x, y over Z2 , and R4 = A2 /I, where I is the
ideal of A2 generated by x3 , y 3 , yx, x2 − xy, y 2 − xy. Identify x and y with their
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H.-L. JIN, H. K. KIM, AND Y. LEE
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images in R4 for simplicity. Then N∗ (R4 ) = J(R4 ) = (x, y), where (x, y) is
the ideal of R4 generated by x and y. So R4 /N∗ (R4 ) ∼
= Z2 , and hence R4 is a
noncommutative tpr ring by Lemma 1.2(1).
(5) Following Xu and Xue [32, Example 7], let A3 = Z4 hx, yi be the free
algebra with noncommuting indeterminates x, y over Z4 , and R5 = A3 /I, where
I is the ideal of A3 generated by x3 , y 2 , yx, x2 −xy, x2 −2, 2x, 2y. Identify x and
y with their images in R5 for simplicity. Then N∗ (R5 ) = J(R5 ) = 2Z4 + (x, y),
where (x, y) is the ideal of R5 generated by x and y. So R5 /N∗ (R5 ) ∼
= Z4 , not
reduced, and hence R5 is not a tpr ring.
The rings R1 , R2 , and R4 in Example 1.9 are minimal noncommutative tpr
rings as we see in the following.
fP
Theorem 1.10. If R is a noncommutative tpr ring of minimal order, then R
is of order 16 and is isomorphic to one of the rings R1 , R2 , and R4 in Example
1.9.
Proof. Let R be a tpr ring of minimal order. Then R is Abelian by Lemma
1.2(3), and any noncommutative Abelian ring of minimal order is isomorphic
to one of the rings in Example 1.9 by [25, Theorem 3.3]. But R1 , R2 , and R4
in Example 1.9 are tpr, so this completes the proof.
Ah
ea
do
The preceding example and theorem imply that a noncommutative Abelian
ring of minimal order need not be tpr.
Let R be a ring and u ∈ R. Recall that u is said to be right regular in R
if ur = 0 implies r = 0 for r ∈ R. The case of left regular is defined similarly;
and u is said to be regular if it is both left and right regular (i.e., not a zero
divisor). Following Lambek [28], an element a of R is called strongly nilpotent
provided that every sequence a0 , a1 , a2 , . . ., such that a0 = a and an+1 ∈ an Ran
for n = 0, 1, 2, . . ., is ultimately zero. N∗ (R) is the set of all strongly nilpotent
elements by [28, Proposition 3.2.1].
Proposition 1.11. Let R be a ring and M be a multiplicatively closed subset
of R consisting of central regular elements. Then R is tpr if and only if so is
RM −1 .
Proof. It suffices to show the necessity by Lemma 1.2(6). Let R be a tpr ring.
Every element of RM −1 can be expressed by au−1 for some a ∈ R and u ∈ M .
This fact is necessary to show that
N (RM −1 ) = N (R)M −1 and N∗ (RM −1 ) = N∗ (R)M −1 .
Clearly N (R)M −1 ⊆ N (RM −1 ). Let au−1 ∈ N (RM −1 ), (au−1 )n = 0
say. Then 0 = (au−1 )n = an u−n implies an = 0 and so a ∈ N (R), entailing
N (M −1 R) = N (R)M −1 .
Now we show that N∗ (RM −1 ) = N∗ (R)M −1 . In fact,
N∗ (RM −1 ) ⊆ N (RM −1 ) = N (R)M −1 = N∗ (R)M −1
by Lemma 1.2(4) because R is tpr, and so N∗ (RM −1 ) ⊆ N∗ (R)M −1 .
RING STRUCTURES CONCERNING FACTORIZATION MODULO RADICALS
9
d2 = d1 e1 f1−1 d1 = (c1 v0−2 f0−1 )e1 f1−1 (c1 v0−2 f0 )−1
2
= c1 e1 c1 v0−2 f0−2 f1−1 (say c2 = c1 e1 c1 ),
...,
dn+1 = dn en fn−1 dn
n
= (cn v0−2 f0−2
n−1
n+1
n
n
n−1
· · · fn−1 )en fn−1 (cn v0−2 f0−2
n−1
f0−2 f1−2
· · · fn−2 fn−1 ),
n−2
f1−2
· · · fn−1 )
fP
= (cn en cn )(v0−2
n−2
f1−2
r in
t
Conversely let cv −1 ∈ N∗ (R)M −1 with c ∈ N∗ (R) and v ∈ M . Write
c0 = c, v0 = v, and d0 = c0 v0−1 . Consider a sequence d0 , d1 = d0 e0 f0−1 d0 =
c0 e0 c0 v0−2 f0−1 (c1 = c0 e0 c0 say),
Ah
ea
do
where ei fi−1 ’s are taken arbitrarily in RM −1 . But c ∈ N∗ (R), so ch eh ch = 0
for some h ≥ 1. This implies that dh = 0, so c0 v0−1 ∈ N∗ (RM −1 ). Therefore
N∗ (RM −1 ) = N∗ (R)M −1 .
Consider now (au−1 )(bv −1 ) ∈ N∗ (RM −1 ) for au−1 , bv −1 ∈ RM −1 . Then
(ab)(uv)−1 ∈ N∗ (RM −1 ). But ab ∈ R and N∗ (RM −1 ) = N∗ (R)M −1 , forcing
ab ∈ N∗ (R). Since R is tpr, R/N∗ (R) is a domain by Lemma 1.2(1); hence we
have a ∈ N∗ (R) or b ∈ N∗ (R). This yields that au−1 ∈ N∗ (RM −1 ) or bv −1 ∈
N∗ (RM −1 ), i.e., the factor ring RM −1 /N∗ (RM −1 ) is a domain. Therefore
RM −1 is tpr by Lemma 1.2(1).
In the proof of Proposition 1.11, we see
N∗ (RM −1 ) = N∗ (R)M −1 = N (R)M −1 = N (RM −1 ).
Let R be a ring. Following the literature, the ring of Laurent
Pnpolynomials,
with an indeterminate x over R, consists of all formal sums i=k ai xi with
canonical addition and multiplication, where ai ∈ R and k, n are (possibly
negative) integers with k ≤ n. This ring is written by R[x; x−1 ], and called the
Laurent polynomial ring over R.
Corollary 1.12. A ring R is tpr if and only if R[x] is tpr if and only if
R[x; x−1 ] is tpr.
Proof. Let R be a ring and M = {1, x, x2 , . . .}. Then, clearly, M is a multiplicatively closed subset of R[x] consisting of central regular elements, and note
R[x; x−1 ] = R[x]M −1 . The result follows Proposition 1.5(1) and Proposition
1.11.
Given any tpr ring we can construct tpr rings, via factor rings of the polynomial ring over it.
Proposition 1.13. Let R be a tpr ring. Then R[x]/(xn ) is a tpr ring for all
n ≥ 1, where (xn ) is the ideal of R[x] generated by xn .
10
H.-L. JIN, H. K. KIM, AND Y. LEE
2. Structure of tunr rings
r in
t
Proof. Write E = R[x]/(xn ). Then N (E) = N∗ (E) = N∗ (R) + x̄E because
(x̄E)n = 0, where we express E by R + x̄E. So E/N∗ (E) ∼
= R/N∗ (R) is a
domain by Lemma 1.2(1) because R is tpr. Thus E is a tpr ring.
In this section we study the structure of rings in which the factor rings
modulo upper nilradicals are domains. A ring R (not necessarily with identity)
will be called tunr if it satisfies the condition:
fP
ai ∈ N ∗ (R) or bi ∈ N ∗ (R) for each i whenever f (x)g(x) ∈ N ∗ (R)[x]
Pm
Pn
for f (x) = i=0 ai xi , g(x) = j=0 bj xj ∈ R[x], where we let m = n by using
zero coefficients if necessary.
Lemma 2.1. (1) A ring R is tunr if and only if R/N ∗ (R) is a domain.
(2) Every tpr ring is tunr.
(3) Let R be a domain and E0 = Dm (R) for m ≥ 1. Then Dn (E0 ) is a tunr
ring for all n ≥ 1.
(4) Let R be a tunr ring. Then Ie (R) = {0, 1}, i.e., I(R) is empty.
(5) If R is a tunr ring, then N (R) = N ∗ (R).
(6) The class of tunr rings is closed under subrings.
Ah
ea
do
Proof. (1) The proof is almost same as one of Lemma 1.2(1).
(2) Let R be a tpr ring. Then N∗ (R) = N ∗ (R) by help of Lemma 1.2(4)
because N∗ (R) ⊆ N ( R), implying that R/N ∗ (R) is a domain. Thus R is tunr
by (1).
The proof of (3) (resp., (4)) is almost same as one of Lemma 1.2(2) (resp.,
Lemma 1.2(3)).
(5) is shown by (1).
(6) Let R be a tunr ring and S be a subring of R. Then N (R) = N ∗ (R)
by (5), so we get N ∗ (S) = N ∗ (R) ∩ S from N ∗ (R) ∩ S ⊆ N ∗ (S) ⊆ N (S) =
N (R) ∩ S = N ∗ (R) ∩ S. The remainder of the proof is almost same as one of
Lemma 1.2(6).
One may ask whether tunr rings are tpr, considering Lemma 2.1(2). However
the following answers negatively.
Example 2.2. We use the ring R in Example 1.3(1). Recall N ∗ (R) = {(aij ) ∈
R | aii = 0 for all i}, so R/N ∗ (R) ∼
= A is a domain. Thus R is tunr by Lemma
2.1(1). But R is not tpr by the argument in Example 1.3.
We see two sorts of conditions under which tunr rings are tpr as follows.
Proposition 2.3. Let R be a ring of bounded index of nilpotency. Then R is
tunr if and only if R is tpr.
Proof. It suffices to show the necessity. Let R be tunr. Then N ∗ (R) = N (R)
by Lemma 2.1(5). By hypothesis, R is of bounded index of nilpotency, and
RING STRUCTURES CONCERNING FACTORIZATION MODULO RADICALS
11
t
so N∗ (R) = N ∗ (R) by the proof of [19, Proposition 1.4]. Then R/N ∗ (R) is a
domain by help of Lemma 2.1(1), and so R is tpr by Lemma 1.2(1).
r in
The ring R in Example 1.3 is not of bounded index of nilpotency.
The following proposition shows that the APR property is a bridge between
tpr and tunr.
Proposition 2.4. Let R be a tunr ring.
(1) If R is an Armendariz ring, then RaR is nilpotent for all a ∈ N (R) (i.e.,
N0 (R) = N∗ (R) = N ∗ (R) = N (R)) and R/N0 (R) is a domain.
(2) If R is a right Goldie ring, then N (R) is nilpotent and R/N0 (R) is a
domain (hence, R is tpr and N0 (R) = N∗ (R) = N ∗ (R) = N (R)).
(3) If R is an APR ring, then R is tpr.
Ah
ea
do
fP
Proof. (1) Let R be an Armendariz ring. Then N0 (R) = N∗ (R) = N ∗ (R) =
N (R) by Lemma 2.1(5) and [22, Lemma 2.3(5)]; hence R/N0 (R) is a domain
by Lemma 2.1(1), and RaR is nilpotent for all a ∈ N (R).
(2) Let R be a right Goldie ring. Then N ∗ (R) is nilpotent by [29], so
N0 (R) = N∗ (R) = N ∗ (R) = N (R) by Lemma 2.1(5). Moreover R/N0 (R) is a
domain by Lemma 2.1(1), so R is tpr by Lemma 1.2(1).
(3) Let R be an APR ring. Then N∗ (R) = N ∗ (R) by [12, Lemma 1.1(7)],
and N (R) = N ∗ (R) by Lemma 2.1(5) because R is tunr, entailing N (R) =
N∗ (R) = N ∗ (R). Hence R/N∗ (R) = R/N ∗ (R) is a domain by Lemma 2.1(5),
so R is tpr by Lemma 1.2(1).
The condition of R being APR in Proposition 2.4(3) is not superfluous.
Indeed, letting R be the semiprime ring in Example 1.3(1), R is not APR by
help of [23, Example 3]. Moreover, E12 ∈ N (R), but RE12 R is not nilpotent
because N0 (R) = N∗ (R) = 0. But R is tunr by Example 2.2.
Considering Proposition 1.5(1), it is natural to ask whether the tunr property
passes to polynomial rings. We answer this question negatively by help of
Smoktunowicz [31].
Example 2.5. Smoktunowicz showed by [31, Corollary 13] that there exists
a nil ring R0 such that polynomial ring over R0 is not nil. Smoktunowicz
constructed the ring R0 so that R0 is an algebra over a countable field K, in
[31, Theorem 12].
Let R be the Dorroh extension of R0 by K. Then clearly N (R) = N ∗ (R) =
R0 . So R/N ∗ (R) ∼
= K, showing that R is tunr by Lemma 2.1(1). Note that
∼ R
∼
N ∗ (R[x]) ⊆ N (R[x]) ⊆ N ∗ (R)[x] = R0 [x] because N ∗R[x]
(R)[x] = N ∗ (R) [x] = K[x].
∗
But R0 [x] is not nil by [31, Corollary 13], entailing N (R[x]) ( N (R[x]).
Thus R[x] is not tunr by Lemma 2.1(5).
If a tunr ring R is of bounded index of nilpotency, then R is tpr by Proposition 2.3, so R[x] is tpr (hence tunr) by Proposition 1.5(1). The ring R in
Example 2.5 is not of bounded index of nilpotency by the construction in the
proof of [31, Theorem 12].
12
H.-L. JIN, H. K. KIM, AND Y. LEE
t
As an extension of tunr rings, we obtain a similar result to Proposition 1.11.
r in
Proposition 2.6. Let R be a ring and M be a multiplicatively closed subset
of R consisting of central regular elements. Then R is tunr if and only if so is
RM −1 .
fP
Proof. It suffices to show the necessity by Lemma 2.1(6). Let R be a tunr
ring. N (RM −1 ) = N (R)M −1 by the proof of Proposition 2.6. We use this fact
freely.
N ∗ (R)M −1 is easily shown to be a nil ideal of RM −1 , so N ∗ (R)M −1 ⊆
∗
N (RM −1 ). We next show the converse inclusion. Let au−1 ∈ N ∗ (RM −1 ).
Then a ∈ N (R) and moreover a = au−1 u ∈ N ∗ (RM −1 ), entailing RaR ⊆
N ∗ (RM −1 ). This implies that RaR is a nil ideal of R, and so a ∈ N ∗ (R).
Thus we get N ∗ (R)M −1 ⊇ N ∗ (RM −1 ), obtaining N ∗ (R)M −1 = N ∗ (RM −1 ).
Consider next (au−1 )(bv −1 ) ∈ N ∗ (RM −1 ) for au−1 , bv −1 ∈ RM −1 . Then
(ab)(uv)−1 in contained in N ∗ (R)M −1 , so ab ∈ N ∗ (R). Since R is tunr,
R/N ∗ (R) is a domain by Lemma 2.1(1); hence we have a ∈ N ∗ (R) or b ∈
N ∗ (R). This yields that au−1 ∈ N ∗ (RM −1 ) or bv −1 ∈ N ∗ (RM −1 ), i.e., the
factor ring RM −1 /N ∗ (RM −1 ) is a domain. Therefore RM −1 is tunr by Lemma
2.1(1).
The following is similar to Corollary 1.12.
Ah
ea
do
Corollary 2.7. (1) For a ring R, R[x] is tunr if and only if so is R[x; x−1 ].
(2) Let R be a ring of bounded index of nilpotency. Then R is tunr if and
only if R[x] is tunr if and only if R[x; x−1 ] is tunr if and only if R is tpr if and
only if R[x] is tpr if and only if R[x; x−1 ] is tpr.
Proof. (1) follows Proposition 2.6.
(2) is shown by Proposition 2.3, Corollary 1.12, and Lemma 2.1(2, 6).
Given any tunr ring we can construct tunr rings, via factor rings of the
polynomial ring over it.
Proposition 2.8. Let R be a tunr ring. Then R[x]/(xn ) is a tunr ring for all
n ≥ 1, where (xn ) is the ideal of R[x] generated by xn .
Proof. The proof is almost similar to one of Proposition 1.13.
3. Structure of tjr rings
In this section we study the structure of rings in which the factor rings
modulo Jacobson radicals are domains. A ring R (not necessarily with identity)
will be called tjr if it satisfies the condition:
ai ∈ J(R) or bi ∈ J(R) for each i whenever f (x)g(x) ∈ J(R)[x]
Pm
Pn
for f (x) = i=0 ai xi , g(x) = j=0 bj xj ∈ R[x], where we let m = n by using
zero coefficients if necessary.
RING STRUCTURES CONCERNING FACTORIZATION MODULO RADICALS
13
r in
t
Lemma 3.1. (1) A ring R is tjr if and only if R/J(R) is a domain.
(2) Let R be a tjr ring. Then Ie (R) = {0, 1}, i.e., I(R) is empty.
(3) Let R be a ring such that J(R) is nil. Then R is tunr if and only if R
is tjr.
(4) If a ring R is tjr, then N (R) ⊆ J(R).
Proof. The proof of (1) (resp., (2)) is almost same as one of Lemma 1.2(1)
(resp., Lemma 1.2(3)).
(3) If J(R) is nil, then N ∗ (R) = J(R), so the result follows.
(4) is an immediate consequence of (1).
fP
The concepts of tpr (tunr) and tjr are independent of each other by the
following.
Ah
ea
do
Example 3.2. (1) There exists a tpr ring but not tjr. We use the ring in [14,
Example 3]. Let R0 be the localization of Z at the prime ideal pZ, where p
is an odd prime; and let R be the quaternions over R0 . Then R is clearly a
domain (hence tpr). But J(R) = pR, and R/J(R) is isomorphic to Mat2 (Zp )
by the argument in [11, Exercise 2A]. Mat2 (Zp ) is not Abelian, so R is not tjr
by Lemma 3.1(2).
(2) There exists a tjr ring but not tpr. Let A be a simple domain and consider
the ring R in Example 1.3 over A. Then N ∗ (R) = {(aij ) ∈ R | aii = 0 for all i},
and R/N ∗ (R) ∼
= A is a simple domain. This yields N ∗ (R) = J(R), so R is tjr.
But R is not tpr by the argument in Example 1.3.
(3) There exists a tjr ring but not tunr. We apply the ring construction and
the argument in [16, Example 1.1]. Let F be a field and let V be a infinite
dimensional left F -module with {v1 , v2 , . . . } a basis. For the endomorphism
ring A = EndF (V ), define
A1 = {f ∈ A | rank(f ) < ∞ and f (vi ) = a1 v1 + · · · + ai vi
for i = 1, 2, . . . with aj ∈ F }
and let R be the F -subalgebra of A generated by A1 and 1A . Then
N∗ (R) = N ∗ (R) = N (R)
= {g ∈ A1 | g(vi ) = a1 v1 + · · · + ai−1 vi−1 for i = 1, 2, . . . with ai ∈ F }
by the computation in [16, Example 1.1].
Consider next the subring
S = {f (x) =
∞
X
ai xi ∈ R[[x]] | a0 ∈ F }
i=0
of R[[x]]. Then J(S) = xR[[x]] and S/J(S) ∼
= F , so S is tjr by Lemma 3.1(1).
But S is not tunr as we see in the following.
Let
n−1
2
f (x) = E12 x + (E34 + E56 )x2 + · · · + (Σi=0
−1
E(2n +2i−1)(2n +2i) )xn + · · ·
14
H.-L. JIN, H. K. KIM, AND Y. LEE
n−1
2
g(x) = E23 x + (E45 + E67 )x2 + · · · + (Σi=0
−1
E(2n +2i)(2n +2i+1) )xn + · · ·
t
and
r in
as in [16, Example 1.1]. Then f (x), g(x) ∈ N (S) (indeed, f (x)2 = 0 = g(x)2 ),
but f (x) + g(x) ∈
/ N (S) by the computation in [16, Example 1.1]. This implies
f (x) ∈
/ N ∗ (S) or g(x) ∈
/ N ∗ (S), so N ∗ (S) ( N (S). Therefore S is not tunr by
Lemma 2.1(5).
fP
Considering Lemma 3.1(3), one may ask whether a tjr ring R is tpr if J(R) is
nil. However the answer is negative by the existence of the ring R in Example
3.2(2). Note that J(R) = N ∗ (R) is nil and R is tjr (if and only if tunr), but R
is not tpr.
In the following we see another role of the ring R in Example 3.2(3). Consider
Proposition 1.5(1). Then it is natural to conjecture that the tpr property passes
to power series rings. However we see a counterexample in the following.
Example 3.3. Let R be the ring in Example 3.2(3), and T be the subring of
R generated by N∗ (R) and 1A . Then T /N∗ (T ) ∼
= F , so T is tpr by Lemma
1.2(1).
Consider next T [[x]]. Then the power series f (x), g(x) ∈ R[[x]] are also contained in N (T [[x]]). Recall f (x)+g(x) ∈
/ N (T [[x]]), and so f (x) ∈
/ N∗ (T [[x]]) or
g(x) ∈
/ N∗ (T [[x]]). Thus T [[x]]/N∗ (T [[x]]) is not reduced (hence not a domain),
forcing T [[x]] to be not tpr.
Ah
ea
do
In the following arguments, we see conditions under which the tpr, tunr, and
tjr properties are equivalent.
Proposition 3.4. (1) Let R be a π-regular ring of bounded index of nilpotency.
Then R is tpr if and only if R is tunr if and only if R is tjr.
(2) Let R be a right or left Artinian ring. Then R is tpr if and only if R is
tunr if and only if R is tjr.
(3) If R is a noncommutative tunr ring of minimal order, then R is of order
16 and is isomorphic to one of the rings R1 , R2 , and R4 in Example 1.9.
(4) If R is a noncommutative tjr ring of minimal order, then R is of order
16 and is isomorphic to one of the rings R1 , R2 , and R4 in Example 1.9.
Proof. (1) Recall that π-regular rings are semiprimitive. So we obtain the result
by Proposition 2.3 and Lemma 3.1(3).
(2) It is well-known that if a ring R is right or left Artinian, then J(R) is
nilpotent. So if R is tpr, tunr, or tjr, then J(R) = N ∗ (R) = N∗ (R) = N (R) by
Lemma 1.2(4), Lemma 2.1(5), and Lemma 3.1(4). Then R is clearly of bounded
index of nilpotency. Therefore we obtain the equivalences among the tpr, tunr,
and tjr properties by help of Lemma 1.2(1), Lemma 2.1(1), and Lemma 3.1(1).
(3) and (4) are proved by (2) and Theorem 1.10.
Dn (R), over a division ring R, satisfies the conditions in Proposition 3.4(1,
2) (refer Lemmas 1.2(2) and 2.1(3)). We see next tjr Laurent polynomial rings.
RING STRUCTURES CONCERNING FACTORIZATION MODULO RADICALS
15
r in
t
Proposition 3.5. (1) Let R be a π-regular ring of bounded index of nilpotency.
Then R is tjr if and only if R[x] is tjr if and only if R[x; x−1 ] is tjr if and only
if R is tunr if and only if R[x] is tunr if and only if R[x; x−1 ] is tunr if and
only if R is tpr if and only if R[x] is tpr if and only if R[x; x−1 ] is tpr.
(2) Let R be a right or left Artinian ring. Then R is tjr if and only if R[x]
is tjr if and only if R[x; x−1 ] is tjr if and only if R is tunr if and only if R[x]
is tunr if and only if R[x; x−1 ] is tunr if and only if R is tpr if and only if R[x]
is tpr if and only if R[x; x−1 ] is tpr.
Proof. The proof is done by Proposition 3.4(1, 2) and Corollary 2.7.
fP
Considering the proofs of Propositions 1.11 and 2.6, one may ask whether
J(RM −1 ) = J(R)M −1 for a given ring R such that M is a multiplicatively
closed subset of R consisting of central regular elements. However the answer
is negative by the following.
Example 3.6. Let R be the ring in Example 3.2(1), i.e., R is the quaternions
over R0 , where R0 is the localization of Z at the prime ideal pZ (p is an odd
prime). Recall that R is a commutative domain and J(R) = pR such that
R/J(R) is isomorphic to M at2 (Zp ). Let M = R\{0}. Then RM −1 is the field
of rational numbers, Q say; and J(RM −1 ) = 0 ( Q = pRM −1 = J(R)M −1 .
Ah
ea
do
Note that J(R) in Example 3.6 is not nil.
Given any tjr ring we can construct tjr rings, via factor rings of the polynomial ring over it.
Proposition 3.7. Let R be a tjr ring. Then R[x]/(xn ) is a tjr ring for all
n ≥ 1, where (xn ) is the ideal of R[x] generated by xn .
Proof. The proof is almost similar to one of Proposition 1.13.
The Dorroh extension does a role in constructing examples of tpr, tunr, and
tjr rings.
Proposition 3.8. (1) Let A be a prime radical ring. Then the Dorroh extension A ⊕D Z of A by Z is tpr.
(2) Let A be a upper nilradical ring. Then the Dorroh extension A ⊕D Z of
A by Z is tunr.
(3) Let A be a Jacobson radical ring. Then the Dorroh extension A ⊕D Z of
A by Z is tjr.
Proof. Let A be a prime radical ring and consider the Dorroh extension A⊕D Z,
R say. Then clearly N (R) = N∗ (R) and R/N∗ (R) ∼
= Z. So R is a tpr ring by
Lemma 1.2(1). The proofs of the remainder are similar.
The nilradicals in Example 1.3(1, 2) can be used to construct tpr rings and
tunr rings, via Dorroh extensions as in Proposition 3.8.
Consider next A = Mat
n (K) over a domain K such that K/J(K) is a domain
P
∞
(e.g., Z), and let R1 = { i=0 ai xi ∈ A[[x]] | a0 ∈ K}, i.e., R1 = K + xA[[x]].
16
H.-L. JIN, H. K. KIM, AND Y. LEE
r in
t
∼ K/J(K), so R1 is a tjr ring by
Then J(R1 ) = J(K) + xA[[x]] and R1 /J(R1 ) =
Lemma 3.1(1). Let R2 = Z + xA[[x]]. Then J(R2 ) = A[[x]] and R2 /J(R2 ) ∼
= Z,
so R2 is also a tjr ring. Note that R2 is isomorphic to the Dorroh extension
xA[[x]] ⊕D Z of xA[[x]] by Z.
fP
Acknowledgments. The authors must thank the referee very much for very
careful reading of the manuscript and many valuable suggestions that improved
the paper by much. The first named author was supported by the National
Natural Science Foundation of China(11361063). And the third named author
was supported by Basic Science Research Program through the National Research Foundation of Korea(NRF) funded by the Ministry of Education(NRF2013R1A1A2A10004687).
References
Ah
ea
do
[1] S. A. Amitsur, A general theory of radicals III, Amer. J. Math. 76 (1954), 126–136.
, Radicals of polynomial rings, Canad. J. Math. 8 (1956), 355–361.
[2]
[3] D. Anderson and V. Camillo, Armendariz rings and Gaussian rings, Comm. Algebra
26 (1998), no. 7, 2265–2272.
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Hai-Lan Jin
Department of Mathematics
Yanbian University
Yanji 133002, P. R. China
E-mail address: [email protected]
Hong Kee Kim
Department of Mathematics and RINS
Gyeongsang National University
Jinju 660-701, Korea
E-mail address: [email protected]
Yang Lee
Department of Mathematics Education
Pusan National University
Pusan 609-735, Korea
E-mail address: [email protected]